Question

\(A(2,1,2), B(1,0,0), C(1+\sqrt{3},\sqrt{3},-\sqrt{6})\) are vertices of a triangle. If the length of the median drawn through \(A\) is \(\lambda\sqrt{9-2\sqrt{3}+2\sqrt{6}}\), then \(\lambda=\)

• A. \(4\)
• B. \(3\)
• C. \(2\)
• D. \(1\)

Hint:

To find the length of a median in coordinate geometry, first find the midpoint of the opposite side and then apply the distance formula between the vertex and that midpoint.

Answer 1

The Correct Option is D

Step 1: Find the three side lengths of the triangle.
With \(A(2,1,2)\), \(B(1,0,0)\), \(C(1+\sqrt3,\sqrt3,-\sqrt6)\):
\[ AB^2 = (2-1)^2+(1-0)^2+(2-0)^2 = 1+1+4 = 6 \]
\[ BC^2 = (\sqrt3)^2+(\sqrt3)^2+(-\sqrt6)^2 = 3+3+6 = 12 \]
\[ AC^2 = (1-\sqrt3)^2+(1-\sqrt3)^2+(2+\sqrt6)^2 = 2(4-2\sqrt3)+(10+4\sqrt6) = 18-4\sqrt3+4\sqrt6 \]

Step 2: Use the standard median-length formula.
For a median drawn from vertex \(A\) to the midpoint of \(BC\), the length satisfies
\[ AM^2 = \frac{2AB^2+2AC^2-BC^2}{4} \]

Step 3: Substitute the values.
\[ AM^2 = \frac{2(6)+2(18-4\sqrt3+4\sqrt6)-12}{4} = \frac{12+36-8\sqrt3+8\sqrt6-12}{4} = \frac{36-8\sqrt3+8\sqrt6}{4} \]

Step 4: Simplify.
\[ AM^2 = 9-2\sqrt3+2\sqrt6 \implies AM = \sqrt{9-2\sqrt3+2\sqrt6} \]

Step 5: Compare with the given form.
Since the given length is \(\lambda\sqrt{9-2\sqrt3+2\sqrt6}\) and we found \(AM = 1\cdot\sqrt{9-2\sqrt3+2\sqrt6}\),
\[ \boxed{\lambda = 1} \]